Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
signed main() {
  if (!false) {
    cin.tie(0);
    ios::sync_with_stdio(0);
  }
  int N, M;
  scanf("%d%d", &N, &M);
  vector<vector<int>> g(N, vector<int>(M));
  vector<int> m(N, INT_MAX), k(M, INT_MAX);
  for (int _i = (N), i = (0); i < _i; i++)
    for (int _j = (M), j = (0); j < _j; j++) {
      scanf("%d", &g[i][j]);
      m[i] = min(m[i], g[i][j]);
      k[j] = min(k[j], g[i][j]);
    }
  for (int _i = (N), i = (1); i < _i; i++)
    for (int _j = (M), j = (1); j < _j; j++)
      if (g[0][0] - g[0][j] != g[i][0] - g[i][j]) {
        printf("-1\n");
        return 0;
      }
  int ans = 0, kin = 0;
  for (int _i = (N), i = (0); i < _i; i++)
    for (int _j = (m[i]), j = (0); j < _j; j++) ++ans;
  for (int _i = (M), i = (0); i < _i; i++)
    for (int _j = (g[0][i] - m[0]), j = (0); j < _j; j++) ++ans;
  for (int _i = (M), i = (0); i < _i; i++)
    for (int _j = (k[i]), j = (0); j < _j; j++) ++kin;
  for (int _i = (N), i = (0); i < _i; i++)
    for (int _j = (g[i][0] - k[0]), j = (0); j < _j; j++) ++kin;
  if (ans < kin) {
    printf("%d\n", ans);
    for (int _i = (N), i = (0); i < _i; i++)
      for (int _j = (m[i]), j = (0); j < _j; j++) printf("row %d\n", i + 1);
    for (int _i = (M), i = (0); i < _i; i++)
      for (int _j = (g[0][i] - m[0]), j = (0); j < _j; j++)
        printf("col %d\n", i + 1);
  } else {
    printf("%d\n", kin);
    for (int _i = (M), i = (0); i < _i; i++)
      for (int _j = (k[i]), j = (0); j < _j; j++) printf("col %d\n", i + 1);
    for (int _i = (N), i = (0); i < _i; i++)
      for (int _j = (g[i][0] - k[0]), j = (0); j < _j; j++)
        printf("row %d\n", i + 1);
  }
  return 0;
}